What mass of glucose can be obtained from starch weighing 200 grams, containing 5% impurities

What mass of glucose can be obtained from starch weighing 200 grams, containing 5% impurities, if the mass fraction of the output is 95%?

Given:
m tech. ((C6H12O5) n) = 200 g
ω (approx.) = 5%
η (C6H12O6) = 95%

To find:
m pract. (C6H12O6) -?

1) (C6H12O5) n + H2O => nC6H12O6;
2) ω ((C6H12O5) n) = 100% – ω (approx.) = 100% – 5% = 95%;
3) m clean. ((C6H12O5) n) = ω * m tech. / 100% = 95% * 200/100% = 190 g;
4) n ((C6H12O5) n) = m pure. / M = 190 / 164n = (1.16 / n) mol;
5) n theory. (C6H12O6) = n ((C6H12O5) n) * n = (1.16 / n) * n = 1.16 mol;
6) n practical (C6H12O6) = η * n theory. / 100% = 95% * 1.16 / 100% = 1.1 mol;
7) m practical. (C6H12O6) = n practical. * M = 1.1 * 180 = 198 g.

Answer: The practical mass of C6H12O6 is 198 g.